NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism

NCERT Solutions for Class 12 Physics Chapter 4 – Free PDF Download

The NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism is given here for the benefit of CBSE Class 12 Science students. In Chapter 4 NCERT Solutions for Class 12 Physics, students are given knowledge on the concepts of Moving Charges and Magnetism. Also, this chapter contains several topics with good marks allotment like magnetic field and direction of a circular coil, magnetic force, etc. are covered.

The concepts present in this chapter of NCERT Solutions for Class 12 Physics are not only important for the CBSE board exam but also are crucial for competitive exams like JEE and NEET. The solutions are created by subject experts following the latest CBSE syllabus and guidelines. Students can now download the PDF of the NCERT Solutions for Class 12 Physics Chapter 4 from the link enclosed below.

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Class 12 Physics NCERT Solutions Moving Charges and Magnetism Important Questions


Q 4.1) A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Answer 4.1:

Given:

The number of turns on the coil (n) is 100

The radius of each turn (r) is  8 cm or 0.08 m

The magnitude of the current flowing in the coil (I) is 0.4 A

The magnitude of the magnetic field at the centre of the coil can be obtained by the following relation:

Bˉ=μ0  2πnI4πr|\bar B| = \frac{\mu_{0}\; 2\pi nI}{4\pi r}

where μ0\mu_{0} is  the permeability of free space = 4π×107  T  m  A14\pi \times 10^{-7}\; T\;m\;A^{-1}

hence,

Bˉ=4π×1074π×2π×100×0.40.08|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\pi \times 100\times 0.4}{0.08}

= 3.14×104  T3.14 \times 10^{-4}\; T

The magnitude of the magnetic field is 3.14×104  T3.14 \times 10^{-4}\; T.

 

Q 4.2) A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Answer 4.2:

The magnitude of the current flowing in the wire (I) is 35 A

The distance of the point from the wire (r) is 20 cm or 0.2 m

At this point, the magnitude of the magnetic field is given by the relation:

Bˉ=μ0  2I4π  r|\bar B| = \frac{\mu_{0}\; 2I}{4\pi\; r}

where,

μ0\mu_{0} = Permeability of free space

= 4π×107  T  m  A14\pi \times 10^{-7}\; T\;m\;A^{-1}

Substituting the values in the equation, we get

Bˉ=4π×1074π×2×350.2|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 35}{0.2}

= 3.5×105  T3.5 \times 10^{-5}\; T

Hence, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5×105  T3.5 \times 10^{-5}\; T.

 

Q 4.3) A long straight wire in the horizontal plane carries a current of 50 A in the north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

Answer 4.3:

The magnitude of the current flowing in the wire is (I) = 50 A.

The point B is 2.5 m away from the East of the wire.

Therefore, the magnitude of the distance of the point from the wire (r) is 2.5 m

The magnitude of the magnetic field at that point is given by the relation:

Bˉ=μ0  2I4π  r|\bar B| = \frac{\mu_{0}\; 2I}{4\pi\; r}

where,

μ0\mu_{0} = Permeability of free space

 

= 4π×107  T  m  A14\pi \times 10^{-7}\; T\;m\;A^{-1}

 

Bˉ=4π×1074π×2×502.5|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 50}{2.5}

 

= 4×106  T4 \times 10^{-6}\; T

The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to Maxwell’s right-hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

 

Q 4.4) A horizontal overhead power line carries a current of 90 A in the east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Answer 4.4:

The magnitude of the current in the power line is (I) = 90 A

The point is located below the electrical cable at distance (r) = 1.5 m

Hence, magnetic field at that point can be calculated as follows,

Bˉ=μ0  2I4π  r|\bar B| = \frac{\mu_{0}\; 2I}{4\pi\; r}

where,

μ0\mu_{0} = Permeability of free space

= 4π×107  T  m  A14\pi \times 10^{-7}\; T\;m\;A^{-1}

Substituting values in the above equation, we get

Bˉ=4π×1074π×2×901.5|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 90}{1.5}

= 1.2×105  T1.2 \times 10^{-5}\; T

The current flows from East to West. The point is below the electrical cable.

Hence, according to Maxwell’s right-hand thumb rule, the direction of the magnetic field is towards the South.

Q 4.5) What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?

Answer 4.5:

In the problem,

The current flowing in the wire is (I) = 8 A

The magnitude of the uniform magnetic field (B) is 0.15 T

The angle between the wire and the magnetic field, θ=30°\theta = 30°.

The magnetic force per unit length on the wire is given as F = BIsinθBI sin\theta

= 0.15×8×1×sin30°0.15\times 8\times 1 \times sin 30°

= 0.6  N  m10.6\; N\; m^{-1}

Hence, the magnetic force per unit length on the wire is 0.6  N  m10.6\; N\; m^{-1}.

 

Q 4.6) A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Answer 4.6:

In the problem,

the length of the wire (l) is 3 cm or 0.03 m

the magnitude of the current flowing in the wire (I) is 10 A

the strength of the magnetic field (B) is 0.27 T

the angle between the current and the magnetic field is θ=90°\theta = 90°.

the magnetic force exerted on the wire is calculated  as follows:

F=BIlsinθF = BIl sin \theta

Substituting the values in the above equation, we get

= 0.27×10×0.03×sin90°0.27\times 10\times 0.03\times sin90°

= 8.1×102  N8.1\times 10^{-2}\; N

Hence, the magnetic force on the wire is 8.1×102  N8.1\times 10^{-2}\; N. The direction of the force can be obtained from Fleming’s left-hand rule.

 

Q 4.7) Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Answer 4.7:

The magnitude of the current flowing in the wire A (IAI_{A}) is 8 A

The magnitude of the current flowing in wire B (IBI_{B}) is 5 A

The distance between the two wires (r) is 4 cm or 0.04 m

The length of the section of wire A (L) = 10 cm = 0.1 m

The force exerted on the length L due to the magnetic field is calculated as follows:

F=μoIAIBL2πrF = \frac{\mu_{o}I_{A}I_{B}L}{2\pi r}

where,

μ0\mu_{0} = Permeability of free space = 4π×107  T  m  A14\pi \times 10^{-7}\; T\;m\;A^{-1}

Substituting the values, we get

F=4π×107×8×5×0.12π×0.04=2×105  NF = \frac{4\pi \times 10^{-7}\times 8\times 5\times 0.1}{2\pi\times 0.04} = 2\times 10^{-5} \;N

The magnitude of force is 2×105  N2\times 10^{-5} \;N. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.

 

Q 4.8) A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

Answer 4.8:

Solenoid length (l) = 80 cm = 0.8 m

Five layers of windings of 400 turn each on the solenoid.

Total number of turns on the solenoid, N = 5 x 400 = 2000

Solenoid Diameter (D) = 1.8 cm = 0.018 m

The current carried by the solenoid (I) = 8.0 A

The relation that gives the magnitude of the magnetic field inside the solenoid near its centre is given below:

B=μ0NIlB = \frac{\mu_{0}NI}{l}

Where,

μ0\mu_{0} = Permeability of free space = 4π×107  T  m  A14\pi \times 10^{-7}\; T\;m\;A^{-1}

 

B=4π×107×2000×80.8B = \frac{4\pi \times 10^{-7}\times 2000\times 8}{0.8}

 

= 2.5×102  T2.5\times 10^{-2}\; T

Hence, the magnitude of B inside the solenoid near its centre is 2.5×102  T2.5\times 10^{-2}\; T.

 

Q 4.9) A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform
the horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Answer 4.9:

In the given problem,

the length of a side of the square coil (l) is 10 cm or 0.1 m

The magnitude of the current flowing in the coil (I) is 12 A

The number of turns on the coil (n) is 20

The angle made by the plane of the coil with B (Magnetic field), θ=30°\theta = 30°

The strength of the magnetic field (B) is 0.8 T

The following relation gives the magnitude of the magnetic torque experienced by the coil in the magnetic field:

τ=n  BIA  sinθ\tau = n\; BIA\; sin\theta

Where,

A = Area of the square coil

= 1 x 1 = 0.1 x 0.1 = 0.01 m2

So, τ=20×0.8×12×0.01×sin30°\tau = 20\times 0.8\times 12\times 0.01\times sin30°

= 0.96 N m

Hence, 0.96 N m is the magnitude of the torque experienced by the coil.

 

Q 4.10) Two moving coil meters, M1 and M2 have the following particulars:

R1=10  Ω,  N1=30R_{1} = 10\; \Omega ,\; N_{1} = 30,

A1=3.6×103  m2,  B1=0.25  TA_{1} = 3.6\times 10^{-3}\; m^{2},\; B_{1} = 0.25\; T

R2=14  Ω,  N2=42R_{2} = 14\; \Omega ,\; N_{2} = 42,

A2=1.8×103  m2,  B2=0.5  TA_{2} = 1.8\times 10^{-3}\; m^{2},\; B_{2} = 0.5\; T

(The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.
Answer 4.10:
Given data:

Moving coil meter M1 Moving coil meter M2
Resistance R1=10  ΩR_{1} = 10\; \Omega R2=14  ΩR_{2} = 14\; \Omega
Number of turns N1=30N_{1} = 30 N2=42N_{2} = 42
Area A1=3.6×103  m2A_{1}=3.6\times 10^{-3}\;m^{2} A2=1.8×103  m2A_{2}=1.8\times 10^{-3}\;m^{2}
Magnetic field strength B1 = 0.25 T B2 = 0.5 T
Spring constant K1 = K K2 = K

(a) Current sensitivity of M1­ is given as:

Is1=N1B1A1K1I_{s1} = \frac{N_{1}B_{1}A_{1}}{K_{1}}

And, Current sensitivity of M2­ is given as:

Is2=N2B2A2K2I_{s2} = \frac{N_{2}B_{2}A_{2}}{K_{2}}

 

RatioIs2Is1=N2B2A2N1B1A1=42×0.5×1.8×103×KK×30×0.25×3.6×103=1.4∴ Ratio \frac{I_{s2}}{I_{s1}} = \frac{N_{2}B_{2}A_{2}}{N_{1}B_{1}A_{1}} = \frac{42\times 0.5\times 1.8\times 10^{-3}\times K}{K\times 30\times 0.25\times 3.6\times 10^{-3}} = 1.4

Hence, the ratio of current sensitivity of M2 to M1 is 1.4.

 

(b) Voltage sensitivity for M2 is given as:

Vs2=N2B2A2K2R2V_{s2} = \frac{N_{2}B_{2}A_{2}}{K_{2}R_{2}}

And, voltage sensitivity for M1 is given as:

Vs1=N1B1A1K1R1V_{s1} = \frac{N_{1}B_{1}A_{1}}{K_{1}R_{1}}

 

RatioIs2Is1=N2B2A2K1R1N1B1A1K2R2=42×0.5×1.8×103×10×KK×14×30×0.25×3.6×103=1∴ Ratio \frac{I_{s2}}{I_{s1}} = \frac{N_{2}B_{2}A_{2}K_{1}R_{1}}{N_{1}B_{1}A_{1}K_{2}R_{2}} = \frac{42\times 0.5\times 1.8\times 10^{-3}\times 10\times K}{K\times 14\times 30\times 0.25\times 3.6\times 10^{-3}} = 1

Hence, the ratio of voltage sensitivity of M2 to M1 is 1.

 

Q 4.11) In a chamber, a uniform magnetic field of 6.5 G (1 G = 10-4 T) is maintained. An electron is shot into the field with a speed of 4.8 x 106 m s-1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e=1.6×1019  C,  me=9.1×1031  kge=1.6\times 10^{-19} \;C,\; m_{e}=9.1\times 10^{-31} \;kg)

Answer 4.11:

Magnetic field strength (B) = 6.5 G = 6.5×104  T6.5\times 10^{-4}\; T

Speed of the electron (v) = 4.8×106  m/s4.8\times 10^{6}\; m/s

Charge on the electron (e) = 1.6×1019  C1.6\times 10^{-19}\; C

Mass of the electron (me) = 9.1×1031  kg9.1\times 10^{-31}\; kg

Angle between the shot electron and magnetic field, θ=90°\theta = 90°

The relation for Magnetic force exerted on the electron in the magnetic field is given as:

F=evB  sinθF = evB\; sin\theta

This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.

Hence, centripetal force exerted on the electron,

Fe=mv2rF_{e} = \frac{mv^{2}}{r}

In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,

Fe = F

=> mv2r=evB  sinθ\frac{mv^{2}}{r} = evB\; sin\theta

=> r=mveB  sinθr = \frac{mv}{eB\;sin\theta}

So,

r=9.1×1031×4.8×1066.5×104×1.6×1019×sin90°=4.2×102  m=4.2cmr = \frac{9.1\times 10^{-31}\times 4.8\times 10^{6}}{6.5\times 10^{-4}\times 1.6\times 10^{-19}\times sin90°} = 4.2\times 10^{-2}\; m = 4.2 cm

Hence, 4.2 cm is the radius of the circular orbit of the electron.

 

Q 4.12) In Exercise 4.11 find the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Answer 4.12:

Magnetic field strength (B) = 6.5×104  T6.5\times 10^{-4}\; T

Charge on the electron (e) = 1.6×1019  C1.6\times 10^{-19}\; C

Mass of the electron (me) = 9.1×1031  kg9.1\times 10^{-31}\; kg

Speed of the electron (v) = 4.8×106  m/s4.8\times 10^{6}\; m/s

Radius of the orbit, r = 4.2 cm = 0.042 m

Frequency of revolution of the electron = v

Angular frequency of the electron = ω=2πv\omega = 2\pi v

Velocity of the electron is related to the angular frequency as: v=rωv = r\omega

In the circular orbit, the magnetic force on the electron is balanced by the centripetal force.

Hence, we can write:

mv2r=evB\frac{mv^{2}}{r} = evB

=> eB=mvr=m(rω)r=m(r.2πv)reB = \frac{mv}{r} = \frac{m(r\omega)}{r} = \frac{m(r.2\pi v)}{r}

=> v=Be2πmv = \frac{Be}{2\pi m}

This expression for frequency is independent of the speed of the electron. On substituting the known values in this expression, we get the frequency as:

v=6.5×104×1.6×10192×3.14×9.1×1031=1.82×106  Hz18  MHzv = \frac{6.5\times 10^{-4}\times 1.6\times 10^{-19}}{2\times 3.14\times 9.1\times 10^{-31}} = 1.82\times 10^{6}\; Hz \approx 18\;MHz

Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.

 

Q 4.13) (a) A circular coil having radius as 8.0 cm, number of turn as 30 and carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 600 with the normal of the coil. To prevent the coil from turning, determine the magnitude of the counter-torque that must be applied.

(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

Answer 4.13:

(a) Number of turns on the circular coil (n) = 30

Radius of the coil (r) = 8.0 cm = 0.08 m

Area of the coil = πr2=π(0.08)2=0.0201  m2\pi r^{2} = \pi (0.08)^{2} = 0.0201 \;m^{2}

Current flowing in the coil (I) = 6.0 A

Magnetic field strength, B = 1 T

The angle between the field lines and normal with the coil surface, θ=60°\theta = 60°

The coil experiences a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the  relation,

τ=nIBA  sinθ\tau = nIBA\; sin\theta

= 30×6×1×0.0201×sin60°30\times 6\times 1\times 0.0201\times sin60°

= 3.133 N m

(b) It can be inferred from the relation τ=nIBA  sinθ\tau = nIBA\; sin\theta that the magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. Hence, the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.

Q 4.14) Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

Answer:

Radius of the coil X, r1 = 16 cm = 0.16 m

Number of turns in coil X, n1 = 20

Current in the coil X, I1 = 16 A

Radius of the coil Y, r2 = 10 cm = 0.1 m

Number of turns in coil Y, n2 = 25

Current in the coil Y, I2 = 18 A

Magnetic field due to the coil X at the centre is given as

B1=μ0N1I12r1B_{1}=\frac{\mu _{0}N_{1}I_{1}}{2r_{1}}

Here, μ0 is the permeability of the free space = 4π x 10-7 Tm/A

B1=4π×107×20×162×0.16B_{1}=\frac{4\pi \times 10^{-7}\times 20\times 16}{2\times 0.16}

= 4π x 10-4 T (Towards east)

Magnetic field due to the coil Y at the centre is given as

B2=μ0N2I22r2B_{2}=\frac{\mu _{0}N_{2}I_{2}}{2r_{2}}

 

B2=4π×107×25×182×0.10B_{2}=\frac{4\pi \times 10^{-7}\times 25\times 18}{2\times 0.10}

 

= 9π x 10-4 T (Towards west)

Therefore, the net magnetic field is given as

B = B2 – B1 = 9π x 10-4 T – 4π x 10-4 T

= 5π x 10-4 T

= 5 x 3.14 x  10-4 

= 1.57 x 10-3 T (Towards west)

Q 4. 15) A magnetic field of 100 G (1 G = 10–4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10–3 m2. The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound around a core is at most 1000 turns m–1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.

Answer:

Magnetic field strength, B = 100 G = 100 x 10-4 T

Number of turns per unit length, N = 1000 turns/m

Current carrying capacity of the coil = 15 A

Permeability of free space, μ0 = 4π x 10-7 TmA-1

Magnetic field is given as

B = μ0 NI/l

⇒ NI/l = B/μ

= (100 x 10-4)/(4π x 10-7)

NI/l = 7961

Now, we can consider a possible combination. Let the current, I = 10 A and the length of the solenoid l = 0.5 m

So we get

(N x 10)/0.5 =  7961

N = 398 turns ≈ 400 turns

Length about 50 cm, number of turns about 400, current about 10 A. These particulars are not unique. Some adjustment with limits is possible

Q 4.16) For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,

B=μ0IR2N2(x2+R2)3/2B=\frac{\mu _{0}IR^{2}N}{2(x^{2}+R^{2})^{3/2}}

(a) Show that this reduces to the familiar result for field at the centre of the coil.
(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,

B=0.72μ0NIRB= 0.72\frac{\mu _{0}NI}{R} , approximately.
[Such an arrangement to produce a nearly uniform magnetic the field over a small region is known as Helmholtz coils.]

Answer:

(a) Given, B=μ0IR2N2(x2+R2)3/2B=\frac{\mu _{0}IR^{2}N}{2(x^{2}+R^{2})^{3/2}}

At the centre of the coil, x = 0

Therefore, the magnetic field at the centre is B=μ0IR2N2R3B=\frac{\mu _{0}IR^{2}N}{2R^{3}}

 

B=μ0IN2RB=\frac{\mu _{0}IN}{2R}

NCERT Class 12 - Chapter 4 - Magnitude of magnetic field- Question 16

(b) Let the small distance between the point P and O be d.

For the coil C, the distance O1P = x1 = (R/2) +d

For the coil d, the distance O2P = x2 = (R/2) – d

The magnetic field on the axis  at a distance x from the centre of the circular coil having a radius a, with n number of turns and having a current I is given by

B=μ04π2πnIa2(x2+a2)3/2B = \frac{\mu _{0}}{4\pi }\frac{2\pi nIa^{2}}{(x^{2}+a^{2})^{3/2}}

 

B=μ02nIa2(x2+a2)3/2B = \frac{\mu _{0}}{2 }\frac{nIa^{2}}{(x^{2}+a^{2})^{3/2}}

The magnetic field at P due to coil C

B1=μ02NIR2(x12+R2)3/2B_{1} = \frac{\mu _{0}}{2 }\frac{NIR^{2}}{(x_{1}^{2}+R^{2})^{3/2}}

 

B1=μ02NIR2((R2+d)2+R2)3/2B_{1} = \frac{\mu _{0}}{2 }\frac{NIR^{2}}{(\left ( \frac{R}{2}+d \right )^{2}+R^{2})^{3/2}}

 

B1=μ02NIR2((R24+d2+Rd)+R2)3/2B_{1} = \frac{\mu _{0}}{2 }\frac{NIR^{2}}{(\left ( \frac{R^{2}}{4}+d^{2}+Rd\right )+R^{2})^{3/2}}

d2 can be neglected when compared to R2

B1=μ0NIR221(5R24)3/2[(1+4d5R)3/2]B_{1} = \frac{\mu _{0}NIR^{2}}{2 }\frac{1}{(\frac{5R^{2}}{4})^{3/2}}\left [ (1+\frac{4d}{5R})^{-3/2} \right ]

It acts along PO2

The magnetic field at P given by

B2=μ02NIa2(x22+R2)3/2B_{2} = \frac{\mu _{0}}{2 }\frac{NIa^{2}}{(x_{2}^{2}+R^{2})^{3/2}}

 

B2=μ02NIR2((R2d)2+R2)3/2B_{2} = \frac{\mu _{0}}{2 }\frac{NIR^{2}}{(\left ( \frac{R}{2}- d \right )^{2}+R^{2})^{3/2}}

 

B2=μ0NIR221(5R24)3/2[(14d5R)3/2]B_{2} = \frac{\mu _{0}NIR^{2}}{2 }\frac{1}{(\frac{5R^{2}}{4})^{3/2}}\left [ (1-\frac{4d}{5R})^{-3/2} \right ]

It acts along PO2

The total Magnetic field at the point P due to the current through the coils B = B1 + B2

B=μ0NIR221(5R24)3/2[(1+4d5R)3/2+(14d5R)3/2]B = \frac{\mu _{0}NIR^{2}}{2 }\frac{1}{(\frac{5R^{2}}{4})^{3/2}}\left [ (1+\frac{4d}{5R})^{-3/2} + (1-\frac{4d}{5R})^{-3/2} \right ]

 

B=μ0NIR22(5R24)3/2[16d5R+1+6d5R]B = \frac{\mu _{0}NIR^{2}}{2\left ( \frac{5R^{2}}{4} \right )^{3/2}}\left [ 1-\frac{6d}{5R}+1+ \frac{6d}{5R} \right ]

 

B=μ0NIR22(5R24)3/2×2B = \frac{\mu _{0}NIR^{2}}{2\left ( \frac{5R^{2}}{4} \right )^{3/2}}\times 2

 

B=μ0NIR(45)3/2B = \frac{\mu _{0}NI}{R}\left ( \frac{4}{5} \right )^{3/2}

 

B=(0.72)μ0NIRB= (0.72)\frac{\mu _{0}NI}{R}

Q 4.17 ) A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field?
(a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.

Answer:

Inner radius of the core, r1= 25 cm = 0.25 m

Outer radius of the core, r2= 26 cm = 0.26 m

Number of turns of the wire, N= 3500 turns

Current in the wire, I = 11 A

(a) The magnetic field outside the toroid is zero.

(b) Inside the core of the toroid, the magnetic field induction is

B = μ0NI/l

Mean length of the toroid,

l=2π(r1+r22)l= 2\pi (\frac{r_{1}+r_{2}}{2})

= π (r1 + r2)  =  π (0.25 + 0.26) = π x 0.51

So, B = μ0NI/

B=(4π×107)×3500×11π×0.51=3.02×102TB = \frac{(4\pi \times 10^{-7})\times 3500\times 11}{\pi \times 0.51} = 3.02 \times 10^{-2} T

(c) The magnetic field in the empty space surrounded by the toroid is zero

Q 4.18) Answer the following questions:
(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say
about the initial velocity of the particle?
(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?
(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in the north to south direction. Specify the direction in which a uniform magnetic field should be set
up to prevent the electron from deflecting from its straight-line path.

Answer:

(a) Initial velocity is either parallel or anti-parallel to the magnetic field. There is no magnetic force acting on the particle when it is parallel or anti-parallel and it moves undeflected.

(b) Yes, because magnetic force can change the direction of velocity but not its magnitude.

(c) Magnetic field should be in a vertically downward direction.

Q 4. 19) An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with a uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field
(a) is transverse to its initial velocity, (b) makes an angle of 30º with the initial velocity.

Answer:

Magnetic field, B = 0.15 T

Potential difference, V = 2.0 kV

An electron gains kinetic energy which is given by

E = (1/2) mv2

eV = (1/2) mv2

v=2eVmv = \sqrt{\frac{2eV}{m}} v=2×1.6×1019×2×1039.1×1031=2.652×107m/sv = \sqrt{\frac{2\times 1.6\times 10^{-19}\times 2\times 10^{3}}{9.1\times 10^{-31}}} = 2.652 \times 10^{7} m/s

(a) When the field applied is transverse to the initial velocity, the force F1 acting on the electron is perpendicular to the direction of the magnetic field and the direction of motion of the electron.  The perpendicular force provides the centripetal force F2 to the electron and makes it move in a circular path.

The forces F1 and F2 are equal

F1 = evB

F2 = mv2/2

F1 = F2

evB = mv2/2

⇒ r = mv/eB

r=9.1×1031×2.652×1070.15×1.6×1019r = \frac{9.1\times 10^{-31}\times 2.652\times 10^{7}}{0.15\times 1.6\times 10^{-19}}

= 10-3 m = 1 mm

The electron will move in a circular path with a radius 1 mm in the direction perpendicular to the magnetic field.

(b) The applied force is inclined to the initial velocity. The velocity of the electron is resolved into two components, V cos θ and V sin θ. V cos θ is along the direction of the field and it causes the electron to move in a straight line. V sin θ is along the normal and it causes the electron to move in a circular path. Therefore, the electron moves in a helical path. The radius of the helical path is given as,

r = mV sin θ/Be

r=9.1×1031×2.652×107×sin3000.15×1.6×1019r = \frac{9.1\times 10^{-31}\times 2.652\times 10^{7}\times sin 30^{0}}{0.15\times 1.6\times 10^{-19}}

= 50.25 x 10-5 m

r = 0.5 mm

Helical trajectory of radius 0.5 mm along the direction of the magnetic field.

Q 4. 20) A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of
the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10–5 V m–1, make a simple guess as to what the beam contains. Why is the answer not unique?

Answer:

Magnetic field, B = 0.75 T

Accelerating voltage, V = 15 kV = 15 x 103 V

Electrostatic field, E = 9.0 × 10–5 V m–1

Kinetic energy of the electron, E = (1/2) mv2

eV = (1/2) mv2

Therefore, (e/m) = (v2/2V)

Here,

Mass of the electron = m

Charge of the electron = e

Velocity of the electron = v

The particles are not deflected by the electric field and the magnetic field. So the force on the particle due to the electric field is balanced by the force due to the magnetic field.

eE = evB

⇒ v = E/B

Therefore, (1/2) m(E/B)2  = eV

e/m = E2/2VB2

=(9.0×105)22×15000×(0.75)2=4.8×107C/kg=\frac{(9.0\times 10^{5})^{2}}{2\times 15000\times (0.75)^{2}}=4.8 \times 10^{7} C/kg

The beam contains Deuterium ions or deuterons

The answer is not unique because only the ratio of charge to mass is determined. Other possible answers are He++, Li++ , etc.

Q 4.21) A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field the same as before?
(Ignore the mass of the wires.) g = 9.8 m s–2.

Answer:

Length of the rod, l = 0.45 m

Mass suspended, m = 60 g = 60 x 10-3 Kg

Current, I = 5 A

(a)  Tension in the wire is zero if the force on the current-carrying wire due to the magnetic field is equal and opposite to the weight on the wire.

BIl = mg

B = mg/Il

=60×103×9.85×0.45=\frac{60\times 10^{-3}\times 9.8}{5\times 0.45} =0. 26133 T

The magnetic field set up normal to the conductor is 0. 26133 T

(b) When the direction of the current is reversed, BIl and mg will act vertically downwards, the effective tension in the wire is

T = BIl+mg

= (0. 26133 x 5.0 x 0.45) + (60 x 10-3 x 9.8) = 0.587 + (0.588) = 1.176 N

Total tension in the wire =  1.176 N

Q 4. 22) The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?

Answer:

Current in the wires, I = 300 A

Distance between the wires, d = 1.5 cm = 0.015 m

Length of the wires, l = 70 cm = 0.7 m

Force between the two wires, F = μ0I2/2πd

Here,

μ= permeability of the free space = 4π x 10-7 Tm/A

F=4π×107×(300)22π×0.015=1.2N/mF =\frac{4\pi \times 10^{-7}\times (300)^{2}}{2\pi \times 0.015}= 1.2 N/m

Since the direction of the wires are opposite, the force will be repulsive.

Q 4. 23)  A uniform magnetic field of 1.5 T exists in a cylindrical region of a radius of 10.0 cm, its direction parallel to the axis along east to west. A wire carrying a current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from N-S to northeast-northwest direction
(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

Answer:

NCERT Class 12 - Chapter 4 - Magnitude of magnetic field- Question 23

(a) Magnetic field , B = 1.5 T

Current in the wire, I = 7.0 A

Radius, r = 10 cm = 0.1 m

Diameter, l = 2 x r = 0.2 m

A force 2.1 N acts vertically downward on the wire

(a) The wire intersects the axis, θ = 900

F = BIlsinθ = 1.5 x 7 x 0.20 x sin 900

= 2.1 N

(b) the wire is turned from N-S to northeast-northwest direction

F = BIl1Sin θ

The length of the wire after turning to northeast-northwest direction is l1 = l/sinθ

Therefore, l = l1 sin θ

F = BIl

= 1.5 x 7 x 0.20= 2.1 N

A force 2.1 N acts vertically downward on the wire

(c) When the wire is lowered by 6 cm, then

Then, x=(10)2(62)=64=8cmx=\sqrt{(10)^{2}-(6^{2})}= \sqrt{64}= 8\, cm

2x = l2 = 16 cm

F2 = BI l2 = 1.5 x 7 x 0.16 = 1. 68 N

The force is directed vertically downwards

Q 4.24) A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in the figure? What is the force on each case? Which case corresponds to stable equilibrium?

NCERT Class 12 - Chapter 4 - Magnitude of magnetic field- Question 24

Answer:

Magnetic field strength, B = 3000 G = 0.3 T

Area of the loop, A = 10 x 5 = 50 cm= 50 x 10-4 m2

Current in the loop, I = 12 A

(a) Torque, τ=IA×B\vec{\tau }=I\vec{A}\times \vec{B}

A is normal to y-z plane and B is directed along the z-axis

A=50×104i^\vec{A}=50\times 10^{-4}\hat{i}

 

B=0.3k^\vec{B}=0.3\hat{k}

 

τ=12×(50×104)i^×0.3k^\vec{\tau }= 12 \times (50\times 10^{-4})\hat{i}\times 0.3\hat{k}

 

=1.8×102j^Nm= -1.8 \times 10^{-2}\hat{j}Nm

The torque is 1.8 x 10-2 N m along the negative y-direction. The net force on the loop is zero since the opposite forces acting on the two ends of the loop will cancel each other.

(b) This is same as (a). Therefore, the torque is 1.8 x 10-2 N m along the negative y-direction. The net force is zero.

(c) A is normal to the x-z plane  and B is along the z-axis

A=50×104j^\vec{A}=- 50\times 10^{-4}\hat{j}

 

B=0.3k^\vec{B}=0.3\hat{k}

 

τ=12×(50×104)j^×0.3k^\vec{\tau }= 12 \times (-50\times 10^{-4})\hat{j}\times 0.3\hat{k}

 

=1.8×102i^Nm= -1.8 \times 10^{-2}\hat{i}Nm

Therefore, the torque is 1.8 x 10-2 N m along the negative x-direction. The net force is zero.

(d) Torque, τ=IA×B\vec{\tau }=I\vec{A}\times \vec{B}

From the given figure, it can be observed that A is normal to the coil. Since the coil makes an angle of -30° with the y axis, A will make an angle 30with the positive x-axis in negative y-direction and B is directed along the z-axis. The angle between A and B is θ = 900. Therefore,

Accordingly the magnitude of torque will be,

τ = 12×50 × 10-4× 0.3

= 1.8×10-2 N m

The direction of torque is (90° + 30°) from negative x -axis or 360°−120° = 240° from positive x-axis. The net force on the loop is zero.

(e)  From the given figure we can see that A is normal to the x-y plane in the positive z-direction and B is directed along the z-axis.

A=50×104k^\vec{A}= 50\times 10^{-4}\hat{k}

 

B=0.3k^\vec{B}= 0.3 \hat{k}

Accordingly,

τ=12×(50×104)k^×0.3k^\vec{\tau }= 12 \times (-50\times 10^{-4})\hat{k}\times 0.3\hat{k}

= 0

Hence, the torque is zero. The force is also zero.

(f) From the given figure we can see that A is normal to the x-y plane in the negative z-direction and B is directed along the z-axis. The angle between A and B is θ = 180°. Therefore,

A=50×104k^\vec{A}= – 50\times 10^{-4}\hat{k}

 

B=0.3k^\vec{B}= 0.3 \hat{k}

 

τ=12×(50×104)k^×0.3k^\vec{\tau }= 12 \times (-50\times 10^{-4})\hat{k}\times 0.3\hat{k}

= 0

Therefore, torque and force is zero.

Case (e) corresponds to stable, and case (f ) corresponds to unstable equilibrium.

Q 4.25) A circular coil of 20 turns and a radius of 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the
(a) total torque on the coil,
(b) the total force on the coil,
(c) average force on each electron in the coil due to the magnetic
field? (The coil is made of copper wire of cross-sectional area 10–5 m2, and
the free electron density in copper is given to be about 1029 m–3.)

Answer:

Number of turns, n = 20 turns

Radius of the coil, r = 10 cm = 0.1 m

Current in the coil, I = 5 A

Magnetic field strength, B = 0.10 T

Cross-sectional area of the wire, a = 10-5 m2

The angle between A\vec{A} and B\vec{B} is θ = 0

(a) total torque on the coil

τ=NIA×B\vec{\tau }=NI\vec{A}\times \vec{B}

⇒ τ = NIAB sin 0 = 0

Therefore, the torque is zero.

(b) Equal and opposite forces are acting on opposite sides of the coil. Therefore, the total force on the coil is zero.

(c) Average force on the moving electron

F = B e vd

vd is the drift velocity of the electrons

vd = I/N e A

Therefore, F = B e I/ N e a

F = B I/n a

0.1×51029×105=5×1025N\frac{0.1\times 5}{10^{29}\times 10^{-5}}= 5\times 10^{-25}N

The average force on each electron is 5 x 10-25 N

Q. 4.26) A solenoid 60 cm long and of radius, 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with an appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? g = 9.8 m s–2.

Answer:

Length of the solenoid, l = 60 cm

Layers of windings = 3

Each layer has 300 turns

Number of turns per unit length, n = (3 x 300)/0.6 = 1500 m-1

Magnetic field inside the solenoid is given by, B = μ0nI

μ0  = 4π x 10-7 T

B =  (4π x 10-7) x 1500 x I

= 6π x 10-4 I———-(1)

Force due to magnetic field, F = I’Bl

I’ = Current in the wire = 6 A

l = length of the wire = 2 cm

The windings of the solenoid would support the weight of the wire when the force due to the magnetic field inside the solenoid balances weight of the wire.

I’Bl = mg

m = mass of the wire = 2.5 g

B = mg/I’l

From equation (1) we get

6π x 10-4  I= mg/I’l

⇒ I = mg/(6π x 10-4  ) I’l

I=2.5×103×9.86×0.02×4π×107×1500=108.37AI = \frac{2.5\times 10^{-3}\times 9.8}{6\times 0.02\times 4\pi \times 10^{-7}\times 1500} = 108.37 A

A current of 108.37 A will support the wire

Q 4.27) A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?

Solution:

Resistance of the galvanometer coil, G = 12 Ω

Current for which there is full scale deflection, I = 3mA

A resistor with a resistance R is connected in series with the galvanometer to convert it into voltmeter. The resistance R is given as

R = (V/Ig) – G

183×10312=600012=5988Ω\frac{18}{3\times 10^{-3}}-12 = 6000 – 12 = 5988 \Omega

A galvanometer can be converted into a voltmeter by connecting a resistor of 5988 Ω

Q 4.28) A galvanometer coil has a resistance of 15 Ω and the metre shows full-scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

Solution:

Resistance of the galvanometer, G = 15 Ω

Current through the galvanometer, Ig = 4 mA = 4 x 10-3 A

Ammeter range = 0 to 6 A

S=IgGIIgS = \frac{I_{g}G}{I-I_{g}}

Therefore, S = S=4×103×1560.004S = \frac{4\times 10^{-3}\times 15}{6-0.004}

= 10 x 10-3 A

S = 10 mA

A 10 mA resistor must be connected in parallel to the galvanometer

Moving charges and magnetism NCERT Class 12 Solutions PDF provided here has answers to textbook questions along with Class 12 Physics Chapter 4 important questions from previous year question papers, exercises and worksheets. These help you in preparing Chapter 4 Physics Class 12 notes, which may be kept handy when the exam nears.

Class 12 Physics NCERT Solutions for Moving Charges and Magnetism

The NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism given here can help the students to clear all their doubts in an instant. These NCERT Solutions of this chapter comprise diagrams, graphs, illustrations and day to day examples that make your study interesting and help you to remember the concepts for a longer period.

The direct derivations of topics like forces between two current-carrying wires, ampere circuital law are frequently asked in exams. Numerical problems are asked from this chapter. The working and figure of cyclotron should be studied intensively.

Subtopics of NCERT Class 12 Physics Chapter 4 Moving Charges and Magnetism

Section Number Topic
4.1 Introduction
4.2 Magnetic Force
4.2.1 Sources And Fields
4.2.2 Magnetic Field, Lorentz Force
4.2.3 Magnetic Force On A current-carrying Conductor
4.3 Motion In A Magnetic Field
4.4 Motion In Combined Electric And Magnetic Fields
4.4.1 Velocity Selector
4.4.2 Cyclotron
4.5 Magnetic Field Due To A Current Element, Biot-savart Law
4.6 Magnetic Field On The Axis Of A Circular Current Loop
4.7 Ampere’s Circuital Law
4.8 The Solenoid And The Toroid
4.8.1 The Solenoid
4.8.2 The Toroid
4.9 Force Between Two Parallel Currents, The Ampere
4.10 Torque On Current Loop, Magnetic Dipole
4.10.1 Torque On A Rectangular Current Loop In A Uniform Magnetic Field
4.10.2 Circular Current Loop As A Magnetic Dipole
4.10.3 The Magnetic Dipole Moment Of A Revolving Electron
4.11 The Moving Coil Galvanometer

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